chemistry of sulphur · 2019-09-30 · sulphur is a yellow solid, insoluble in water, soluble in...
TRANSCRIPT
1
All high school revision materials are available on www.kusoma.co.ke
Chemistry of
SULPHUR
2
All high school revision materials are available on www.kusoma.co.ke
A.SULPHUR (S)
Sulphur is an element in Group VI Group 16)of the Periodic table . It has atomic
number 16 and electronic configuration 16 and valency 2 /divalent and thus
forms the ion S2-
A. Occurrence.
Sulphur mainly occurs :
(i) as free element in Texas and Louisiana in USA and Sicily in Italy.
(ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near
Naivasha in Kenya
(iii)as copper pyrites(CuFeS2) ,Galena (PbS,Zinc blende(ZnS))and iron
pyrites(FeS2) in other parts of the world.
B. Extraction of Sulphur from Fraschs process
Suphur occurs about 200 metres underground. The soil structure in these areas
is usually weak and can easily cave in.
Digging of tunnels is thus discouraged in trying to extract the mineral.
Sulphur is extracted by drilling three concentric /round pipes of diameter of
ratios 2:8: 18 centimeters.
Superheated water at 170oC and 10atmosphere pressure is forced through the
outermost pipe.
The high pressures ensure the water remains as liquid at high temperatures
instead of vapour of vapour /gas.
The superheated water melts the sulphur because the melting point of sulphur is
lower at about at about 115oC.
A compressed air at 15 atmospheres is forced /pumped through the innermost
pipe.
The hot air forces the molten sulphur up the middle pipe where it is collected
and solidifies in a large tank.
It is about 99% pure.
Diagram showing extraction of Sulphur from Fraschs Process
3
All high school revision materials are available on www.kusoma.co.ke
C. Allotropes of Sulphur.
1. Sulphur exist as two crystalline allotropic forms:
(i)Rhombic sulphur
(ii)Monoclinic sulphur
Rhombic sulphur Monoclinic sulphur
Bright yellow crystalline solid
Has a melting point of 113oC
Has a density of 2.06gcm-3
Stable below 96oC
Has octahedral structure
Pale yellow crystalline solid
Has a melting point of 119oC
Has a density of 1.96gcm-3
Stable above 96oC
Has a needle-like structure
Rhombic sulphur and Monoclinic sulphur have a transition temperature of
96oC.This is the temperature at which one allotrope changes to the other.
4
All high school revision materials are available on www.kusoma.co.ke
5
All high school revision materials are available on www.kusoma.co.ke
2. Sulphur exists in non-crystalline forms as:
(i)Plastic sulphur-
Plastic sulphur is prepared from heating powdered sulphur to boil
then pouring a thin continuous stream in a beaker with cold water. A long thin
elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright
yellow crystalline rhombic sulphur.
(ii)Colloidal sulphur-
Colloidal sulphur is formed when sodium thiosulphate (Na2S2O3) is
added hydrochloric acid to form a yellow precipitate.
D. Heating Sulphur.
A molecule of sulphur exists as puckered ring of eight atoms joined by
covalent bonds as S8.
On heating the yellow sulphur powder melts at 113oC to clear amber liquid with
low viscosity and thus flows easily.
On further heating to 160oC the molten liquid darkens to a brown very viscous
liquid that does not flow easily.
This is because the S8 rings break into S8 chain that join together to form very
long chains made of over 100000 atoms of Sulphur.
The long chains entangle each other reducing their mobility /flow and hence
increases their viscosity.
On continued further heating to above 160oC, the viscous liquid darkens but
becomes more mobile/flows easily and thus less viscous.
This is because the long chains break to smaller/shorter chains.
At 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2
molecules that solidifies to S8 ring of “flowers of sulphur” on the cooler parts.
Summary of changes on heating sulphur
6
All high school revision materials are available on www.kusoma.co.ke
Observation on heating Explanation/structure of Sulphur
Solid sulphur
Heat to 113oC
Amber yellow liquid
Heat to 160oC
Liquid darkens
Heat to 444oC
Liquid boils to brown vapour
Cool to room temperature
Yellow sublimate
(Flowers of Sulphur)
Puckered S8 ring
Puckered S8 ring in liquid form (low
viscosity/flow easily)
Puckered S8 ring break/opens then join
to form long chains that entangle (very
high viscosity/very low rate of flow)
Mixture of S8 ,S6 ,S2 vapour
Puckered S8 ring
E. Physical and Chemical properties of Sulphur.(Questions)
1. State three physical properties unique to Sulphur
Sulphur is a yellow solid, insoluble in water, soluble in carbon
disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity.
It has a melting point of 115oC and a boiling point of 444
oC.
2. Moist/damp/wet blue and red litmus papers were put in a gas jar
containing air/oxygen. Burning sulphur was then lowered into the gas jar.
State and explain the observation made.
Observations
-Sulphur melts then burns with a blue flame
Colourless gas produced that has a pungent smell
Red litmus paper remains red. Blue litmus paper turns red.
Explanation
Sulphur burns in air and faster in Oxygen to form Sulphur(IV)Oxide gas
and traces/small amount of Sulphur(VI)Oxide gas. Both oxides react with water
to form the corresponding acidic solution i.e
(i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid
(ii) Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid
7
All high school revision materials are available on www.kusoma.co.ke
Chemical equation
S(s) + O2(g) -> SO2(g) (Sulphur(IV)Oxide gas)
2S(s) + 3O2(g) -> 2SO3(g) (Sulphur(VI)Oxide gas traces)
SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid)
SO3(g) + H2O(l) -> H2 SO4 (aq) ( sulphuric(VI)acid).
3. Iron filings were put in a test tube containing powdered sulphur then
heated on a Bunsen flame. Stop heating when reaction starts. State and
explain the observations made. Test the effects of a magnet on the mixture
before and after heating. Explain.
Observations
Before heating, the magnet attracts iron filings leaving sulphur
After heating, the magnet does not attract the mixture.
After heating, a red glow is observed that continues even when heating is
stopped..
Black solid is formed.
Explanation
Iron is attracted to a magnet because it is ferromagnetic.
When a mixture of iron and sulphur is heated, the reaction is exothermic giving
out heat energy that makes the mixture to continue glowing even after stopping
heating.
Black Iron(II)sulphide is formed which is a compound and thus not
ferromagnetic.
Chemical equation
Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ -∆H)
Heated powdered heavy metals combine with sulphur to form black sulphides.
Cu(s) + S(s) -> CuS(s)
Zn(s) + S(s) -> ZnS(s)
Pb(s) + S(s) -> PbS(s)
4.The set up below show the reaction of sulphur on heated concentrated
sulphuric(VI)acid.
8
All high school revision materials are available on www.kusoma.co.ke
(i)State and explain the observation made.
Observation
Yellow colour of sulphur fades
Orange colour of potassium dichromate(VI)paper turns to green.
Explanation
Hot concentrated sulphuric(VI)acid oxidizes sulphur to sulphur (IV)oxide gas.
The oxide is also reduced to water. Traces of sulphur (VI)oxide is formed.
Chemical equation
S(s) + 3H2 SO4 (l) -> 3SO2(g) + 3H2O(l) +SO3(g)
Sulphur (IV)oxide gas turns Orange potassium dichromate(VI)paper to green.
(ii)State and explain the observation made if concentrated sulphuric (VI)
acid is replaced with concentrated Nitric (V) acid in the above set up.
Observation
Yellow colour of sulphur fades
Colurless solution formed
Brown fumes/gas produced.
Explanation
Hot concentrated Nitric(V)acid oxidizes sulphur to sulphuric (VI)acid. The
Nitric (V) acid is reduced to brown nitrogen(IV)oxide gas.
Chemical equation
S(s) + 6HNO3 (l) -> 6NO2(g) + 2H2O(l) +H2SO4 (l)
NB:
Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur
like the other mineral acids.
5.State three main uses of sulphur .
Sulphur is mainly used in:
9
All high school revision materials are available on www.kusoma.co.ke
(i)Contact process for the manufacture/industrial/large scale production
of concentrated sulphuric(VI)acid.
(ii)Vulcanization of rubber to make it harder, tougher, stronger, and more
durable.
(iii)Making gun powder and match stick heads
(iv) As ointments to treat fungal infections
6. Revision Practice
The diagram below represents the extraction of sulphur by Fraschs
process. Use it to answer the questions that follow.
(a)Name the substances that passes through:
M Superheated water at 170oC and 10 atmosphere pressure
L Hot compressed air
N Molten sulphur
(b)What is the purpose of the substance that passes through L and M?
M- Superheated water at 170oC and 10 atmosphere pressure is used to
melt the sulphur
L- Hot compressed air is used to force up the molten sulphur.
(c) The properties of the two main allotropes of sulphur represented by
letters A and B are given in the table below. Use it to answer the questions
that follow.
A B
Appearance Bright yellow Pale yellow
Density(gcm-3
) 1.93 2.08
Melting point(oC) 119 113
Stability Above 96oC Below 96
oC
I.What are allotropes?
Different forms of the same element existing at the same temperature and
pressure without change of state.
II. Identify allotrope:
A. Monoclinic sulphur
L M
N
10
All high school revision materials are available on www.kusoma.co.ke
B . Rhombic sulphur
III. State two main uses of sulphur.
-Manufacture of sulphuric(VI)acid
-as fungicide
-in vulcanization of rubber to make it harder/tougher/ stronger
-manufacture of dyes /fibres
(d)Calculate the volume of sulphur (IV)oxide produced when 0.4 g of
sulphur is completely burnt in excess air.(S = 32.0 ,I mole of a gas occupies
24 dm3 at room temperature)
Chemical equation
S(s) + O2(g) -> SO2(g)
Mole ratio S: SO2 = 1:1
Method 1 32.0 g of sulphur -> 24 dm3 of SO2(g) 0.4 g of sulphur -> 0.4 g x 24 dm3 = 0.3 dm3 32.0 g
Method 2 Moles of sulphur used = Mass of sulphur => 0.4 = 0.0125 moles
Molar mass of sulphur 32
Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles
Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one
mole of gas =>0.0125 moles x 24 dm3 = 0.3 dm3
11
All high school revision materials are available on www.kusoma.co.ke
B.COMPOUNDS OF SULPHUR The following are the main compounds of sulphur:
(i) Sulphur(IV)oxide
(ii) Sulphur(VI)oxide .
(iii) Sulphuric(VI)acid
(iv) Hydrogen Sulphide
(v) Sulphate(IV)/SO32-
and Sulphate(VI)/ SO42-
salts
(i) Sulphur(IV)oxide(SO2)
(a) Occurrence Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from
geysersand hot springs in active volcanic areas of the world e.g. Olkaria and
Hells gate near Naivasha in Kenya.
(b) School laboratory preparation
In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the
reaction of
Method 1:Using Copper and Sulphuric(VI)acid.
12
All high school revision materials are available on www.kusoma.co.ke
Method 2:Using Sodium Sulphate(IV) and hydrochloric acid.
(c)Properties of Sulphur(IV)oxide(Questions)
13
All high school revision materials are available on www.kusoma.co.ke
1. Write the equations for the reaction for the formation of sulphur
(IV)oxide using:
(i)Method 1
Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l)
Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l)
Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l)
Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l)
Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover
unreacted metals stopping further reaction thus producing very small
amount/quantity of sulphur (IV)oxide gas.
(ii)Method 2
Na2SO3(aq) + HCl(aq) -> NaCl(aq ) + SO2(g) + 2H2O(l)
K2SO3(aq) + HCl(aq) -> KCl(aq ) + SO2(g) + 2H2O(l)
BaSO3(s) + 2HCl(aq) -> BaCl2(aq ) + SO2(g) + H2O(l)
CaSO3(s) + 2HCl(aq) -> CaCl2(aq ) + SO2(g) + H2O(l)
PbSO3(s) + 2HCl(aq) -> PbCl2(s ) + SO2(g) + H2O(l)
Lead(II)chloride is soluble on heating thus reactants should be heated to
prevent it coating/covering unreacted PbSO3(s)
2.State the physical properties unique to sulphur (IV)oxide gas.
Sulphur (IV)oxide gas is a colourless gas with a pungent irritating and choking
smell which liquidifies easily. It is about two times denser than air.
3. The diagram below show the solubility of sulphur (IV)oxide gas. Explain.
14
All high school revision materials are available on www.kusoma.co.ke
Sulphur(IV) oxide is very soluble in water.
One drop of water dissolves all the Sulphur (IV) oxide in the flask leaving a
vacuum.
If the clip is removed, atmospheric pressure forces the water up through the
narrow tube to form a fountain to occupy the vacuum.
An acidic solution of sulphuric (IV)acid is formed which turns litmus solution
red.
Chemical equation
SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid turn litmus red)
4.Dry litmus papers and wet/damp/moist litmus papers were put in a gas
jar containing sulphur(IV) oxide gas. State and explain the observations
made.
Observations
(i)Dry Blue litmus paper remains blue.
Dry red litmus paper remains red.
(ii) Wet/damp/moist blue litmus paper turns red.
Moist/damp/wet red litmus paper remains red.
15
All high school revision materials are available on www.kusoma.co.ke
Both litmus papers are then bleached /decolorized.
Explanation
Dry sulphur(IV) oxide gas is a molecular compound that does not
dissociate/ionize to release H+(aq)ions and thus has no effect on dry blue/red
litmus papers.
Wet/damp/moist litmus papers contain water that dissolves /react with dry
sulphur(IV) oxide gas to form a solution of weak sulphuric(IV)acid (H2 SO3
(aq)).
Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions:
H2 SO3 (aq) -> 2H+(aq) + SO3
2- (aq)
The free H+(aq)ions are responsible for turning blue litmus paper turns red
showing the gas is acidic.
The SO32-
(aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for
many reactions of the gas.
It is easily/readily oxidized to sulphate(VI) SO42-
(aq) ions making sulphur(IV)
oxide gas act as a reducing agent as in the following examples:
(a)Bleaching agent
Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when
put in sulphur(IV) oxide gas.
This is because sulphur(IV) oxide removes atomic oxygen from the coloured
dye/ material to form sulphuric(VI)acid.
Chemical equations
(i)Formation of sulphuric(IV)acid
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Decolorization/bleaching of the dye/removal of atomic oxygen.
Method I. H2 SO3 (aq) + (dye + O) -> H2 SO4 (aq) + dye
(coloured) (colourless)
Method II. H2 SO3 (aq) + (dye) -> H2 SO4 (aq) + (dye - O)
(coloured) (colourless)
Sulphur(IV) oxide gas therefore bleaches by reduction /removing oxygen from a
dye unlike chlorine that bleaches by oxidation /adding oxygen.
The bleaching by removing oxygen from Sulphur(IV) oxide gas is temporary.
This is because the bleached dye regains the atomic oxygen from the
atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its
original colour. e.g.
16
All high school revision materials are available on www.kusoma.co.ke
Old newspapers turn brown on exposure to air on regaining the atomic oxygen.
The bleaching through adding oxygen by chlorine gas is permanent.
(b)Turns Orange acidified potassium dichromate(VI) to green
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified
potassium dichromate(VI) solution. or;
(ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar
containing Sulphur(IV) oxide gas.
Observation:
Orange acidified potassium dichromate(VI) turns to green.
Explanation:
Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange
Cr2O72-
ions to green Cr3+
ions without leaving a residue itself oxidized from
SO32-
ions in sulphuric(IV) acid to SO42-
ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32-
(aq)
(iii)Oxidation of SO32-
(aq)and reduction of Cr2O72-
(aq)
3SO32-
(aq) + Cr2O7
2-(aq) +8H+(aq) -> 3SO4
2-(aq) + 2Cr
3+(aq) + 4H2O(l)
This is a confirmatory test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also reduces acidified potassium dichromate(VI) from
orange Cr2O72-
ions to green Cr3+
ions leaving a yellow residue.
(c)Decolorizes acidified potassium manganate(VII)
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified
potassium manganate(VII) solution. or;
(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar
containing Sulphur(IV) oxide gas.
17
All high school revision materials are available on www.kusoma.co.ke
Observation:
Purple acidified potassium manganate(VII) turns to colourless/ acidified
potassium manganate(VII) is decolorized.
Explanation:
Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple
MnO4- ions to green Mn
2+ ions without leaving a residue itself oxidized from
SO32-
ions in sulphuric(IV) acid to SO42-
ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32-
(aq)
(iii)Oxidation of SO32-
(aq)and reduction of MnO4- (aq)
5SO32-
(aq) + 2MnO4
- (aq) +6H+(aq) -> 5SO4
2-(aq) + 2Mn
2+(aq) + 3H2O(l)
(purple) (colourless)
This is another test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from
purple MnO4- ions to colourless Mn
2+ ions leaving a yellow residue.
(d)Decolorizes bromine water
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine
water . or;
(ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide
gas. Swirl.
Observation:
Yellow bromine water turns to colourless/ bromine water is decolorized.
Explanation:
Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic
acid (HBr) without leaving a residue itself oxidized from SO32-
ions in sulphuric
(IV) acid to SO42-
ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
18
All high school revision materials are available on www.kusoma.co.ke
H2 SO3 (aq) -> 2H+(aq) + SO32-
(aq)
(iii)Oxidation of SO32-
(aq)and reduction of MnO4- (aq)
SO32-
(aq) + Br2 (aq) + H2O(l) -> SO4
2-(aq) + 2HBr(aq)
(yellow) (colourless)
This can also be used as another test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving
a yellow residue.
(e)Reduces Iron(III) Fe3+
salts to Iron(II) salts Fe2+
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3
of Iron (III)chloride solution. or;
(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing
Sulphur(IV) oxide gas.Swirl.
Observation:
Yellow/brown Iron (III)chloride solution turns to green
Explanation:
Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown
Fe3+
ions to green Fe2+
ions without leaving a residue itself oxidized from SO32-
ions in sulphuric(IV) acid to SO42-
ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32-
(aq)
(iii)Oxidation of SO32-
(aq)and reduction of Fe3+
(aq)
SO32-
(aq) + 2Fe
3+ (aq) +3H2O(l) -> SO4
2-(aq) + 2Fe
2+(aq) + 2H
+(aq)
(yellow) (green)
(f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3
of concentrated nitric(V)acid. or;
(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing
Sulphur(IV) oxide gas. Swirl.
19
All high school revision materials are available on www.kusoma.co.ke
Observation:
Brown fumes of a gas evolved/produced.
Explanation:
Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown
nitrogen(IV)oxide gas itself oxidized from SO32-
ions in sulphuric(IV) acid to
SO42-
ions in sulphuric(VI) acid.
Chemical/ionic equation:
SO2(g) + 2HNO3 (l) -> H2 SO4 (l) + NO2 (g)
(brown fumes/gas)
(g)Reduces Hydrogen peroxide to water
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3
of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or
Barium chloride followed by five drops of 2M hydrochloric acid/ 2M nitric(V)
acid.
Observation:
A white precipitate is formed that persist /remains on adding 2M hydrochloric
acid/ 2M nitric(V) acid.
Explanation:
Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself
oxidized from SO32-
ions in sulphuric(IV) acid to SO42-
ions in sulphuric(VI)
acid.
When Ba2+
ions in Barium Nitrate(V) or Barium chloride solution is added, a
white precipitate of insoluble Barium salts is formed showing the presence of of
either SO32-
,SO42-
,CO32-
ions. i.e.
Chemical/ionic equation:
SO32-
(aq) + Ba
2+ (aq) -> BaSO3(s)
white precipitate
SO42-
(aq) + Ba
2+ (aq) -> BaSO4(s)
white precipitate
CO32-
(aq) + Ba
2+ (aq) -> BaCO3(s)
white precipitate
If nitric(V)/hydrochloric acid is added to the three suspected insoluble white
precipitates above, the white precipitate:
(i) persist/remains if SO42-
(aq)ions (BaSO4(s)) is present.
(ii)dissolves if SO32-
(aq)ions (BaSO3(s)) and CO32-
(aq)ions (BaCO3(s))is
present. This is because:
20
All high school revision materials are available on www.kusoma.co.ke
I. BaSO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO2
gas that turns Orange moist filter paper dipped in acidified Potassium
dichromate to green.
Chemical equation
BaSO3(s) +2H+(aq)
-> Ba
2+ (aq) + SO2(g) + H2O(l)
I. BaCO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic
CO2 gas that forms a white precipitate when bubbled in lime water.
Chemical equation
BaCO3(s) +2H+(aq)
-> Ba
2+ (aq) + CO2(g) + H2O(l)
5.Sulphur(IV)oxide also act as an oxidizing agent as in the following
examples.
(a)Reduction by burning Magnesium
Experiment
Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide
gas
Observation
Magnesium ribbon continues to burn with difficulty.
White ash and yellow powder/speck
Explanation
Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to
produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and
oxygen.
The metal continues to burn on Oxygen forming white Magnesium oxide
solid/ash.
Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.
During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the
metal is oxidized (reducing agent)
Chemical equation
SO2(g) + 2Mg(s) -> 2MgO(s) + S(s)
(white ash/solid) (yellow speck/powder)
(b)Reduction by Hydrogen sulphide gas
Experiment
Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas
Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas.
21
All high school revision materials are available on www.kusoma.co.ke
Or
Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas
Invert a gas jar full of hydrogen sulphide gas over the gas jar containing
Sulphur(IV)oxide gas. Swirl
Observation
Yellow powder/speck
Explanation
Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur
residue and itself reduced to also sulphur that form on the sides of reaction
flask/gas jar.
A little moisture/water act as catalyst /speeds up the reaction.
Chemical equation
SO2(g) + 2H2S(g) -> 2H2O(l) + 3S(s)
(yellow speck/powder)
6.Sulphur(IV)oxide has many industrial uses. State three.
(i)In the contact process for the manufacture of Sulphuric(VI)acid
(ii)As a bleaching agent of pulp and paper.
(iii)As a fungicide to kill microbes’
(iv)As a preservative of jam, juices to prevent fermentation
(ii) Sulphur(VI)oxide(SO3)
(a) Occurrence
Sulphur (VI)oxide is does not occur free in nature/atmosphere
(b) Preparation
In a Chemistry school laboratory Sulphur (VI)oxide may prepared from:
Method 1;Catalytic oxidation of sulphur(IV)oxide gas.
Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed
through Concentrated Sulphuric(VI)acid .
The dry mixture is then passed through platinised asbestos to catalyse/speed up
the combination to form Sulphur (VI)oxide gas.
Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a
freezing mixture /ice cold water.
The solid fumes out on heating to a highly acidic poisonous gas.
Chemical equation 2SO2(g) + O2(g) --platinised asbestos--> 2SO3 (g)
Method 2; Heating Iron(II)sulphate(VI) heptahydrate
22
All high school revision materials are available on www.kusoma.co.ke
When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a
boiling tube ,it loses the water of crystallization and colour changes from green
to white.
Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
(green solid) (white solid)
On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes
to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas.
Sulphur (VI) oxide readily / easily solidify as white silky needles when the
mixture is passed through a freezing mixture/ice cold water.
Iron(III)oxide is left as a brown residue/solid.
Chemical equation
2FeSO4 (s) -> Fe2O3(s) + SO2 (g) + SO3(g)
(green solid) (brown solid)
Caution
On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous
fumes of concentrated sulphuric(VI)acid and thus its preparation in a school
laboratory is very risky.
(c) Uses of sulphur(VI)oxide
One of the main uses of sulphur(VI)oxide gas is as an intermediate product in
the contact process for industrial/manufacture/large scale/production of
sulphuric(VI)acid.
(iii) Sulphuric(VI)acid(H2SO4)
(a) Occurrence
Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three
mineral acids;
Nitric(V)acid
Sulphuric(VI)acid
Hydrochloric acid.
Mineral acids do not occur naturally but are prepared in a school laboratory and
manufactured at industrial level.
(b)The Contact process for industrial manufacture of H2SO4 .
I. Raw materials
23
All high school revision materials are available on www.kusoma.co.ke
The main raw materials for industrial preparation of Sulphuric(VI)acid include:
(i)Sulphur from Fraschs process or from heating metal sulphide ore like
Galena(PbS),Zinc blende(ZnS)
(ii)Oxygen from fractional distillation of air
(iii)Water from rivers/lakes
II. Chemical processes
The contact process involves four main chemical processes:
(i)Production of Sulphur (IV)oxide
As one of the raw materials, Sulphur (IV)oxide gas is got from the following
sources;
I. Burning/roasting sulphur in air.
Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide
gas in the burners
Chemical equation
S(s) + O2(g) --> SO2 (g)
II. Burning/roasting sulphide ores in air.
Sulphur (IV)oxide gas is produced as a by product in extraction of some metals
like:
- Lead from Lead(II)sulphide/Galena,(PbS)
- Zinc from zinc(II)sulphide/Zinc blende, (ZnS)
- Copper from Copper iron sulphide/Copper pyrites, (CuFeS2)
On roasting/burning, large amount /quantity of sulphur(IV)oxide is
generated/produced.
Chemical equation
(i)2PbS (s) + 3O2 (g) -> 2PbO(s) + 2SO2 (g)
(ii)2ZnS (s) + 3O2 (g) -> 2ZnO(s) + 2SO2 (g)
(ii)2CuFeS2 (s) + 4O2 (g) -> 2FeO(s) + 3SO2 (g) + Cu2O(s)
Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far
distance safely.
(ii)Purification of Sulphur(IV)oxide
Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities.
These impurities “poison”/impair the catalyst by adhering on/covering its
surface.
24
All high school revision materials are available on www.kusoma.co.ke
The impurities are removed by electrostatic precipitation method .
In the contact process Platinum or Vanadium(V)oxide may be used.
Vanadium(V)oxide is preferred because it is :
(i) cheaper/less expensive
(ii) less easily poisoned by impurities
(iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide
Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC
in a heat exchanger.
The heated mixture is passed through long pipes coated with pellets of
Vanadium (V)oxide catalyst.
The close “contact” between the reacting gases and catalyst give the process its
name.
Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to
Sulphur(VI)oxide gas.
Chemical equation
2SO2 (g) + O2(g) -- V2O5 --> 2SO2 (g)
This reaction is exothermic (-∆H) and the temperatures need to be maintained at
around 450oC to ensure that:
(i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not
too slow/long at lower temperatures below 450oC
(ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide
gas and Oxygen gas at higher temperatures than 450oC.
(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid
Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid.
Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of
very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is
highly exothermic.
To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of
98% Sulphuric(VI)acid in the absorption chamber/tower.
The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric
(VI) acid/ pyrosulphuric (VI) acid.
Chemical equation
H2SO4 (aq) + SO3 (g) -> H2S2O7 (l)
Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully
with distilled water to give concentrated sulphuric (VI) acid .
Chemical equation
H2S2O7 (l) + H2O (l) -> 2H2SO4 (l)
The acid is stored ready for market/sale.
25
All high school revision materials are available on www.kusoma.co.ke
III. Environmental effects of contact process
Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that
form acid rain if they escape to the atmosphere.
In the Contact process, about 2% of these gases do not form sulphuric (VI)
acid.
The following precautions prevent/minimize pollution from Contact process:
(i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat
exchangers.
(ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid
instead of water.
This prevents the formation of fine droplets of the corrosive/
toxic/poisonous fumes of concentrated sulphuric (VI) acid.
(iii)scrubbing-This involves passing the exhaust gases through very tall
chimneys lined with quicklime/calcium hydroxide solid.
This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate
(IV) /CaSO3
Chemical equation
Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g)
IV. Uses of Sulphuric(VI)acid
Sulphuric (VI) acid is used:
(i) in making dyes and paint
(ii)as acid in Lead-acid accumulator/battery
(iii) for making soapless detergents
(iv) for making sulphate agricultural fertilizers
VI. Sketch chart diagram showing the Contact process
26
All high school revision materials are available on www.kusoma.co.ke
(c) Properties of Concentrated sulphuric(VI)acid
(i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of
1.84gcm-3
.It has a boiling point of 338oC.
(ii) Concentrated sulphuric(VI)acid is very soluble in water.
The solubility /dissolution of the acid very highly exothermic.
The concentrated acid should thus be diluted slowly in excess water.
Water should never be added to the acid because the hot acid scatters highly
corrosive fumes out of the container.
(iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+
ions.
Free H+ ions are responsible for turning the blue litmus paper red. Concentrated
sulphuric (VI) acid thus do not change the blue litmus paper red.
(iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the
atmosphere and do not form a solution.
This makes concentrated sulphuric (VI) acid very suitable as drying agent
during preparation of gases.
(v)The following are some chemical properties of concentrated sulphuric (VI)
acid:
I. As a dehydrating agent
Experiment I;
Put about four spatula end fulls of brown sugar and glucose in separate 10cm3
beaker.
Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand
for about 10 minutes.
Observation;
Colour( in brown sugar )change from brown to black.
Colour (in glucose) change from white to black.
10cm3 beaker becomes very hot.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically and physically combined elements of water(Hydrogen
and Oxygen in ratio 2:1)from compounds.
When added to sugar /glucose a vigorous reaction that is highly exothermic take
place.
The sugar/glucose is charred to black mass of carbon because the acid
dehydrates the sugar/glucose leaving carbon.
Caution
This reaction is highly exothermic that start slowly but produce fine particles of
carbon that if inhaled cause quick suffocation by blocking the lung villi.
Chemical equation
27
All high school revision materials are available on www.kusoma.co.ke
Glucose: C6H12O6(s) --conc.H2SO4--> 6C (s) + 6H2O(l)
(white) (black)
Sugar: C12H22O11(s) --conc.H2SO4--> 12C (s) +11H2O(l)
(brown) (black)
Experiment II;
Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a
boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid
.Warm .
Observation;
Colour change from blue to white.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes
physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from
hydrated compounds.
The acid dehydrates blue copper(II)sulphate to white anhydrous
copper(II)sulphate .
Chemical equation
CuSO4.5H2O(s) --conc.H2SO4--> CuSO4 (s) + 5H2O(l)
(blue) (white)
Experiment III;
Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about
10cm3 of concentrated sulphuric (VI) acid.
Place moist/damp/wet filter paper dipped in acidified potassium
dichromate(VI)solution on the mouth of the boiling tube. Heat strongly.
Caution:
Absolute ethanol is highly flammable.
Observation;
Colourless gas produced.
Orange acidified potassium dichromate (VI) paper turns to green.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically combined elements of water(Hydrogen and Oxygen in
ratio 2:1)from compounds.
The acid dehydrates ethanol to ethene gas at about 170oC.
Ethene with =C=C= double bond turns orange acidified potassium dichromate
(VI) paper turns to green.
Chemical equation
28
All high school revision materials are available on www.kusoma.co.ke
C2H5OH(l) --conc.H2SO4/170oC --> C2H4 (g) + H2O(l)
NB: This reaction is used for the school laboratory preparation of ethene gas
Experiment IV;
Put about 4cm3 of methanoic acid in a boiling tube .Carefully add about 6 cm3
of concentrated sulphuric (VI) acid. Heat gently
Caution: This should be done in a fume chamber/open
Observation;
Colourless gas produced.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes
chemically combined elements of water (Hydrogen and Oxygen in ratio
2:1)from compounds.
The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas.
Chemical equation
HCOOH(l) --conc.H2SO4 --> CO(g) + H2O(l)
NB: This reaction is used for the school laboratory preparation of small amount
carbon (II)oxide gas
Experiment V;
Put about 4cm3 of ethan-1,2-dioic/oxalic acid in a boiling tube .Carefully add
about 6 cm3 of concentrated sulphuric (VI) acid. Pass any gaseous product
through lime water.Heat gently
Caution: This should be done in a fume chamber/open
Observation;
Colourless gas produced.
Gas produced forms a white precipitate with lime water.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically combined elements of water (Hydrogen and Oxygen in
ratio 2:1)from compounds.
The acid dehydrates ethan-1,2-dioic/oxalic acid to a mixture of
poisonous/toxic carbon(II)oxide and carbon(IV)oxide gases.
Chemical equation
HOOCCOOH(l) --conc.H2SO4 --> CO(g) + CO2(g) + H2O(l)
NB: This reaction is also used for the school laboratory preparation of small
amount carbon (II) oxide gas.
29
All high school revision materials are available on www.kusoma.co.ke
Carbon (IV) oxide gas is removed by passing the mixture through concentrated
sodium/potassium hydroxide solution.
II. As an Oxidizing agent
Experiment I
Put about 2cm3 of Concentrated sulphuric (VI) acid into three separate boiling
tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium
dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of
Copper turnings, Zinc granule and Iron filings to each boiling tube separately.
Observation;
Effervescence/fizzing/bubbles
Blue solution formed with copper,
Green solution formed with Iron
Colourless solution formed with Zinc
Colourless gas produced that has a pungent irritating choking smell.
Gas produced turn orange moist/damp/wet filter paper dipped in acidified
potassium dichromate (VI)solution to green.
Explanation
Concentrated sulphuric (VI) acid is strong oxidizing agent.
It oxidizes metals to metallic sulphate(VI) salts and itself reduced to
sulphur(IV)oxide gas.
Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in
acidified potassium dichromate (VI)solution to green.
CuSO4(aq) is a blue solution. ZnSO4(aq) is a colourless solution. FeSO4(aq) is a
green solution.
Chemical equation
Cu(s) + 2H2SO4(aq) --> CuSO4(aq) + SO2(g) + 2H2O(l)
Zn(s) + 2H2SO4(aq) --> ZnSO4(aq) + SO2(g) + 2H2O(l)
Fe(s) + 2H2SO4(aq) --> FeSO4(aq) + SO2(g) + 2H2O(l)
Experiment II
Put about 2cm3 of Concentrated sulphuric (VI) acid into two separate boiling
tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium
dichromate (VI)solution on the mouth of the boiling tube.
Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube
separately.
Warm.
Observation;
Black solid charcoal dissolves/decrease
Yellow solid sulphur dissolves/decrease
30
All high school revision materials are available on www.kusoma.co.ke
Colourless gas produced that has a pungent irritating choking smell.
Gas produced turn orange moist/damp/wet filter paper dipped in acidified
potassium dichromate (VI)solution to green.
Explanation
Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes non-
metals to non metallic oxides and itself reduced to sulphur(IV)oxide gas.
Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in
acidified potassium dichromate (VI)solution to green.
Charcoal is oxidized to carbon(IV)oxide. Sulphur is oxidized to
Sulphur(IV)oxide .
Chemical equation
C(s) + 2H2SO4(aq) --> CO2(aq) + 2SO2(g) + 2H2O(l)
S(s) + 2H2SO4(aq) --> 3SO2(g) + 2H2O(l)
III. As the least volatile acid
Study the table below showing a comparison in boiling points of the three
mineral acids
Mineral acid Relative molecula mass Boiling point(oC)
Hydrochloric acid(HCl) 36.5 35.0
Nitric(V)acid(HNO3) 63.0 83.0
Sulphuric(VI)acid(H2SO4) 98.0 333
1.Which is the least volatile acid? Explain
Sulphuric(VI)acid(H2SO4) because it has the largest molecule and joined by
Hydrogen bonds making it to have the highest boiling point/least volatile.
2. Using chemical equations, explain how sulphuric(VI)acid displaces the
less volatile mineral acids.
(i)Chemical equation
KNO3(s) + H2SO4(aq) --> KHSO4(l) + HNO3(g)
NaNO3(s) + H2SO4(aq) --> NaHSO4(l) + HNO3(g)
This reaction is used in the school laboratory preparation of Nitric(V) acid
(HNO3).
(ii)Chemical equation
KCl(s) + H2SO4(aq) --> KHSO4(s) + HCl(g)
NaCl(s) + H2SO4(aq) --> NaHSO4(s) + HCl(g)
This reaction is used in the school laboratory preparation of Hydrochloric acid
(HCl).
(d) Properties of dilute sulphuric(VI)acid.
31
All high school revision materials are available on www.kusoma.co.ke
Dilute sulphuric(VI)acid is made when about 10cm3 of concentrated sulphuric
(VI) acid is carefully added to about 90cm3 of distilled water.
Diluting concentrated sulphuric (VI) acid should be done carefully because the
reaction is highly exothermic.
Diluting concentrated sulphuric (VI) acid decreases the number of moles
present in a given volume of solution which makes the acid less corrosive.
On diluting concentrated sulphuric(VI) acid, water ionizes /dissociates the acid
fully/wholly into two(dibasic)free H+(aq) and SO4
2-(aq)ions:
H2SO4 (aq) -> 2H+(aq) + SO4
2-(aq)
The presence of free H+(aq)ions is responsible for ;
(i)turn litmus red because of the presence of free H+(aq)ions
(ii)have pH 1/2/3 because of the presence of many free H+(aq)ions hence
a strongly acidic solution.
(iii)Reaction with metals
Experiment:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test
tubes. Add about 0.1g of Magnesium ribbon to one test tube. Cover the mixture
with a finger as stopper. Introduce a burning splint on top of the finger and
release the finger “stopper”. Repeat by adding Zinc, Copper and Iron instead of
the Magnesium ribbon.
Observation:
No effervescence/ bubbles/ fizzing with copper
Effervescence/ bubbles/ fizzing with Iron ,Zinc and Magnesium
Colourless gas produced that extinguishes burning splint with a “pop”
sound.
Colourless solution formed with Zinc and Magnesium.
Green solution formed with Iron
Explanation:
When a metal higher than hydrogen in the reactivity/electrochemical
series is put in a test tube containing dilute sulphuric(VI)acid,
effervescence/ bubbling/ fizzing takes place with evolution of Hydrogen
gas.
Impure hydrogen gas extinguishes burning splint with a “pop” sound.
A sulphate (VI) salts is formed. Iron, Zinc and Magnesium are higher
than hydrogen in the reactivity/electrochemical series.
They form Iron (II)sulphate(VI), Magnesium sulphate(VI) and Zinc
sulphate(VI).
. When a metal lower than hydrogen in the reactivity/electrochemical
series is put in a test tube containing dilute sulphuric(VI)acid, there is no
effervescence/ bubbling/ fizzing that take place.
32
All high school revision materials are available on www.kusoma.co.ke
Copper thus do not react with dilute sulphuric(VI)acid.
Chemical/ionic equation
Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g)
Mg(s) + 2H+(aq) --> Mg
2+ (aq) + H2(g)
Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g)
Zn(s) + 2H+(aq) --> Zn
2+ (aq) + H2(g)
Fe(s) + H2SO4(aq) --> FeSO4(aq) + H2(g)
Fe(s) + H+(aq) --> Fe
2+ (aq) + H2(g)
NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that
cover/coat the unreacted metals.
(ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid
(iv)Reaction with metal carbonates and hydrogen carbonates
Experiment:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean
boiling tubes. Add about 0.1g of sodium carbonate to one boiling tube.
Introduce a burning splint on top of the boiling tube. Repeat by adding Zinc
carbonate, Copper (II)carbonate and Iron(II)Carbonate in place of the sodium
hydrogen carbonate.
Observation:
Effervescence/ bubbles/ fizzing.
Colourless gas produced that extinguishes burning splint.
Colourless solution formed with Zinc carbonate, sodium hydrogen
carbonate and sodium carbonate.
Green solution formed with Iron(II)Carbonate
Blue solution formed with Copper(II)Carbonate
Explanation:
When a metal carbonate or a hydrogen carbonates is put in a test tube
containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing
takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas
extinguishes a burning splint and forms a white precipitate when bubbled
in lime water.
A sulphate (VI) salts is formed.
Chemical/ionic equation
ZnCO3(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) + CO2(g)
ZnCO3(s) + 2H+(aq) --> Zn
2+ (aq) + H2O(l) + CO2(g)
CuCO3(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l) + CO2(g)
CuCO3(s) + 2H+(aq) --> Cu
2+ (aq) + H2O(l) + CO2(g)
FeCO3(s) + H2SO4(aq) --> FeSO4(aq) + H2O(l) + CO2(g)
33
All high school revision materials are available on www.kusoma.co.ke
FeCO3(s) + 2H+(aq) --> Fe
2+ (aq) + H2O(l) + CO2(g)
2NaHCO3(s) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l) + 2CO2(g)
NaHCO3(s) + H+(aq) --> Na
+ (aq) + H2O(l) + CO2(g)
Na2CO3(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l) + CO2(g)
NaHCO3(s) + H+(aq) --> Na
+ (aq) + H2O(l) + CO2(g)
(NH4)2CO3(s) + H2SO4(aq) --> (NH4)2SO4 (aq) + H2O(l) + CO2(g)
(NH4)2CO3 (s) + H+(aq) --> NH4
+ (aq) + H2O(l) + CO2(g)
2NH4HCO3(aq) + H2SO4(aq) --> (NH4)2SO4 (aq) + H2O(l) + CO2(g)
NH4HCO3(aq) + H+(aq) --> NH4
+ (aq) + H2O(l) + CO2(g)
NB:
Calcium, Lead and Barium carbonates forms insoluble sulphate(VI)salts that
cover/coat the unreacted metals.
(v)Neutralization-reaction of metal oxides and alkalis/bases
Experiment I:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean
boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir.
Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)Oxide in place
of the Copper(II)Oxide.
Observation:
Blue solution formed with Copper(II)Oxide
Colourless solution formed with other oxides
Explanation:
When a metal oxide is put in a test tube containing dilute
sulphuric(VI)acid, the oxide dissolves forming a sulphate (VI) salt.
Chemical/ionic equation
ZnO(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l)
ZnO(s) + 2H+(aq) --> Zn
2+ (aq) + H2O(l)
CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)
CuO(s) + 2H+(aq) --> Cu
2+ (aq) + H2O(l)
MgO(s) + H2SO4(aq) --> MgSO4(aq) + H2O(l)
MgO(s) + 2H+(aq) --> Mg
2+ (aq) + H2O(l)
Na2O(s) + H2SO4(aq) --> Na2SO4(aq) + H2O(l)
Na2O(s) + 2H+(aq) --> 2Na
+ (aq) + H2O(l)
K2CO3(s) + H2SO4(aq) --> K2SO4(aq) + H2O(l)
34
All high school revision materials are available on www.kusoma.co.ke
K2O(s) + H+(aq) --> 2K
+ (aq) + H2O(l)
NB:
Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that
cover/coat the unreacted metals oxides.
Experiment II:
Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of
0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of
phenolphthalein indicator.Titrate the acid to get a permanent colour
change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium
hydroxide solution
Observation:
Colour of phenolphthalein changes from pink to colourless at the end point.
Explanation
Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a
sulphate salt and water only.
Colour of the indicator used changes when a slight excess of acid is added to
the base at the end point
Chemical equation:
2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + H2O(l)
OH-(s) + H
+(aq) --> H2O(l)
2KOH(aq) + H2SO4(aq) --> K2SO4(aq) + H2O(l)
OH-(s) + H
+(aq) --> H2O(l)
2NH4OH(aq) + H2SO4(aq) --> (NH4)2SO4(aq) + H2O(l)
OH-(s) + H
+(aq) --> H2O(l)
35
All high school revision materials are available on www.kusoma.co.ke
(iv) Hydrogen sulphide(H2S)
(a) Occurrence Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water
from geysers and hot springs in active volcanic areas of the world e.g. Olkaria
and Hells gate near Naivasha in Kenya.
It is present in rotten eggs and human excreta.
(b) Preparation
Hydrogen sulphide is prepared in a school laboratory by heating Iron (II)
sulphide with dilute hydrochloric acid.
(c) Properties of Hydrogen sulphide(Questions) 1. Write the equation for the reaction for the school laboratory preparation
of Hydrogen sulphide.
Chemical equation: FeS (s) + 2HCl (aq) -> H2S (g) FeCl2 (aq)
2. State three physical properties unique to Hydrogen sulphide.
Hydrogen sulphide is a colourless gas with characteristic pungent poisonous
smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is
denser than water and turns blue litmus paper red.
3. Hydrogen sulphide exist as a dibasic acid when dissolved in water. Using
a chemical equation show how it ionizes in aqueous state.
36
All high school revision materials are available on www.kusoma.co.ke
H2S(aq) -> H+(aq) + HS
-(aq)
H2S(aq) -> 2H+(aq) + S
2- (aq)
Hydrogen sulphide therefore can form both normal and acid salts e.g
Sodium hydrogen sulphide and sodium sulphide both exist
4. State and explain one gaseous impurity likely to be present in the gas jar
containing hydrogen sulphide above.
Hydrogen/ H2
Iron(II)sulphide contains Iron as impurity .The iron will react with dilute
hydrochloric acid to form iron(II)chloride and produce hydrogen gas that mixes
with hydrogen sulphide gas.
5. State and explain the observations made when a filter paper dipped in
Lead(II) ethanoate /Lead (II) nitrate(V) solution is put in a gas jar
containing hydrogen sulphide gas.
Observations
Moist Lead(II) ethanoate /Lead (II) nitrate(V) paper turns black.
Explanation
When hydrogen sulphide is bubbled in a metallic salt solution, a metallic
sulphide is formed.
All sulphides are insoluble black salts except sodium sulphide, potassium
sulphide and ammonium sulphides.
Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V)
paper .
The gas reacts with Pb2+
in the paper to form black Lead(II)sulphide.
This is the chemical test for the presence of H2S other than the physical smell of
rotten eggs.
Chemical equations
Pb2+
(aq) + H2S -> PbS + 2H+(aq)
(black)
Fe2+
(aq) + H2S -> FeS + 2H+(aq)
(black)
Zn2+
(aq) + H2S -> ZnS + 2H+(aq)
(black)
Cu2+
(aq) + H2S -> CuS + 2H+(aq)
(black)
2Cu+(aq) + H2S -> Cu2S + 2H+(aq)
(black)
6. Dry hydrogen sulphide was ignited as below.
Flame A
Dry Hydrogen sulphide gas
37
All high school revision materials are available on www.kusoma.co.ke
(i) State the observations made in flame A
Hydrogen sulphide burns in excess air with a blue flame to form
sulphur(IV)oxide gas and water.
Chemical equation: 2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g)
Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid
and water.
Chemical equation: 2H2S(g) + O2(g) -> 2H2O(l) + 2S(s)
7. Hydrogen sulphide is a strong reducing agent that is oxidized to yellow
solid sulphur as precipitate. The following experiments illustrate the
reducing properties of Hydrogen sulphide.
(a)Turns Orange acidified potassium dichromate(VI) to green
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified
potassium dichromate (VI) solution. or;
(ii)Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas
jar containing Hydrogen sulphide gas.
Observation:
Orange acidified potassium dichromate (VI) turns to green.
Yellow solid residue.
Explanation:
Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange
Cr2O72-
ions to green Cr3+
ions leaving a yellow solid residue as itself is
oxidized to sulphur.
Chemical/ionic equation:
4H2S(aq) + Cr2O7
2-(aq) +6H+(aq) -> 4S(aq) + 2Cr
3+(aq) + 7H2O(l)
This test is used for differentiating Hydrogen sulphide and sulphur (IV)oxide
gas.
Sulphur(IV)oxide also reduces acidified potassium dichromate(VI) from orange
Cr2O72-
ions to green Cr3+
ions without leaving a yellow residue.
(b)Decolorizes acidified potassium manganate(VII)
Experiment:
38
All high school revision materials are available on www.kusoma.co.ke
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified
potassium manganate(VII) solution. or;
(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar
containing Hydrogen Sulphide gas.
Observation:
Purple acidified potassium manganate(VII) turns to colourless/ acidified
potassium manganate(VII) is decolorized.
Yellow solid residue.
Explanation:
Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple
MnO4- ions to green Mn
2+ ions leaving a residue as the gas itself is oxidized to
sulphur.
Chemical/ionic equation:
5H2S(g) + 2MnO4
- (aq) +6H+(aq) -> 5S (s) + 2Mn
2+(aq) + 8H2O(l)
(purple) (colourless)
This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide
gas.
Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from
purple MnO4- ions to colourless Mn
2+ ions leaving no yellow residue.
(c)Decolorizes bromine water
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine
water . or;
(ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide
gas. Swirl.
Observation:
Yellow bromine water turns to colourless/ bromine water is decolorized.
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces yellow bromine water to colourless
hydrobromic acid (HBr) leaving a yellow residue as the gas itself is oxidized to
sulphur.
Chemical/ionic equation:
H2 S(g)
+ Br2 (aq) -> S (s) + 2HBr(aq)
(yellow solution) (yellow solid) (colourless)
This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide
gas.
39
All high school revision materials are available on www.kusoma.co.ke
Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from
purple MnO4- ions to colourless Mn
2+ ions leaving no yellow residue.
(d)Reduces Iron(III) Fe3+
salts to Iron(II) salts Fe2+
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3
of Iron (III)chloride solution. or;
(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing
Hydrogen sulphide gas. Swirl.
Observation:
Yellow/brown Iron (III)chloride solution turns to green.
Yellow solid solid
Explanation:
Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown
Fe3+
ions to green Fe2+
ions leaving a yellow residue.The gas is itself oxidized
to sulphur.
Chemical/ionic equation:
H2S(aq)
+ 2Fe3+
(aq) -> S (s) + Fe2+
(aq) + 2H+(aq)
(yellow solution) (yellow residue) (green)
(e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3
of concentrated nitric(V)acid. or;
(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing
Hydrogen sulphide gas. Swirl.
Observation:
Brown fumes of a gas evolved/produced.
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown
nitrogen(IV)oxide gas itself oxidized to yellow sulphur.
Chemical/ionic equation:
H2S(g) + 2HNO3 (l) -> 2H2O(l) + S (s) + 2NO2 (g)
(yellow residue) (brown fumes)
(f)Reduces sulphuric(VI)acid to Sulphur
Experiment:
40
All high school revision materials are available on www.kusoma.co.ke
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3
of concentrated sulphuric(VI)acid. or;
(ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar
containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow
sulphur.
Chemical/ionic equation:
3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s)
(yellow residue)
(g)Reduces Hydrogen peroxide to water
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3
of 20 volume hydrogen peroxide.
Observation:
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and
itself oxidized to yellow sulphur
Chemical/ionic equation:
H2S(g) + H2O2 (l) -> 2H2O(l) + S (s)
(yellow residue)
8.Name the salt formed when:
(i)equal volumes of equimolar hydrogen sulphide neutralizes sodium
hydroxide solution:
Sodium hydrogen sulphide
Chemical/ionic equation:
H2S(g) + NaOH (l) -> H2O(l) + NaHS (aq)
(ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide
solution:
Sodium sulphide
Chemical/ionic equation:
H2S(g) + 2NaOH (l) -> 2H2O(l) + Na2S (aq)
Practice
41
All high school revision materials are available on www.kusoma.co.ke
Hydrogen sulphide gas was bubbled into a solution of metallic
nitrate(V)salts as in the flow chart below
(a)Name the black solid Copper(II)sulphide
(b)Identify the cation responsible for the formation of:
I. Blue solution Cu2+
(aq)
II. Green solution Fe2+
(aq)
III. Brown solution Fe3+
(aq)
(c)Using acidified potassium dichromate(VI) describe how you would
differentiate between sulphur(IV)Oxide and hydrogen sulphide
-Bubble the gases in separate test tubes containing acidified Potassium
dichromate(VI) solution.
-Both changes the Orange colour of acidified Potassium dichromate(VI)
solution to green.
-Yellow solid residue/deposit is formed with Hydrogen sulphide
Chemical/ionic equation:
4H2S(aq) + Cr2O7
2-(aq) +6H+(aq) -> 4S(aq) + 2Cr
3+(aq) + 7H2O(l)
3SO32-
(aq) + Cr2O7
2-(aq) +8H+(aq) -> 3SO4
2-(aq) + 2Cr
3+(aq) + 4H2O(l)
(d)State and explain the observations made if a burning splint is introduced
at the mouth of a hydrogen sulphide generator.
ObservationGas continues burning with a blue flame
Explanation: Hydrogen sulphide burns in excess air with a blue flame to
form sulphur(IV)oxide gas and water.
Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g)
Hydrogen sulphide
Blue solution Brown solution
Black solid Green solution
42
All high school revision materials are available on www.kusoma.co.ke
(v)Sulphate (VI) (SO42-
)and Sulphate(IV) (SO32-
) salts 1. Sulphate (VI) (SO4
2-) salts are normal and acid salts derived from Sulphuric
(VI)acid H2SO4.
2. Sulphate(IV) (SO32-
) salts are normal and acid salts derived from Sulphuric
(IV)acid H2SO3.
3. Sulphuric (VI)acid H2SO4 is formed when sulphur(VI)oxide gas is bubbled in
water.
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore
the Sulphate (VI) (SO42-
) and hydrogen sulphate (VI) (HSO4-) salts.
i.e.
H2SO4 (aq) -> 2H+(aq) + SO4
2-(aq)
H2SO4 (aq) -> H+(aq) + HSO4
-(aq)
All Sulphate (VI) (SO42-
) salts dissolve in water/are soluble except Calcium (II)
sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II)
sulphate (VI) (PbSO4)
All Hydrogen sulphate (VI) (HSO3-) salts exist in solution/dissolved in water.
Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate
(VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as
solids.
Other Hydrogen sulphate (VI) (HSO4-) salts do not exist except those of
Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II)
hydrogen sulphate (VI) (Mg (HSO4)2).
4. Sulphuric (IV)acid H2SO3 is formed when sulphur(IV)oxide gas is bubbled in
water.
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore
the Sulphate (IV) (SO32-
) and hydrogen sulphate (VI) (HSO4-) salts.
i.e.
H2SO3 (aq) -> 2H+(aq) + SO3
2-(aq)
H2SO3 (aq) -> H+(aq) + HSO3
-(aq)
43
All high school revision materials are available on www.kusoma.co.ke
All Sulphate (IV) (SO32-
) salts dissolve in water/are soluble except Calcium (II)
sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II)
sulphate (IV) (PbSO3)
All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water.
Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate
(IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as
solids.
Other Hydrogen sulphate (IV) (HSO3-) salts do not exist except those of
Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II)
hydrogen sulphate (IV) (Mg (HSO3)2).
5.The following experiments show the effect of heat on sulphate(VI) (SO42-
)and
sulphate(IV) (SO32-
) salts:
Experiment:
In a clean dry test tube place separately about 1.0g of :
Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I)
sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any
gases produced using litmus papers.
Observations:
-Colourless droplets of liquid forms on the cooler parts of the test tube in all
cases.
-White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate
(VI) and Sodium (I) sulphate (IV).
-Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI)
-Colour changes from blue to white then black in case of Copper (II) sulphate
(VI)
-Blue litmus paper remain and blue and red litmus paper remain red in case of
Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV)
-Blue litmus paper turns red and red litmus paper remain red in case of Iron
(II)sulphate(VI) and Copper (II) sulphate (VI).
Explanation
(i)All Sulphate (VI) (SO42-
) salts exist as hydrated salts with water of
crystallization that condenses and collects on cooler parts of test tube as a
colourless liquid on gentle heating. e.g.
K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l)
Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l)
MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l)
CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l)
ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l)
FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
44
All high school revision materials are available on www.kusoma.co.ke
CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l)
All Sulphate (VI) (SO42-
) salts do not decompose on heating except Iron (II)
sulphate (VI) and Copper (II) sulphate (VI).
(i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic
sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a
brown /yellow residue.
Chemical equation
2FeSO4 (s) -> Fe2O3(s) + SO2(g) + SO3(g)
This reaction is used for the school laboratory preparation of small amount of
sulphur(VI)oxide gas.
Sulphur (VI) oxide readily /easily solidifies as white silky needles when the
mixture is passed through freezing mixture/ice cold water.
Sulphur (IV) oxide does not.
(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II)
oxide and Sulphur (VI) oxide gas.
Chemical equation
2CuSO4 (s) -> CuO(s) + SO3(g)
This reaction is used for the school laboratory preparation of small amount of
sulphur(VI)oxide gas.
6. The following experiments show the test for the presence of sulphate (VI)
(SO42-
)and sulphate(IV) (SO32-
) ions in a sample of a salt/compound:
Experiments/Observations:
(a)Using Lead(II)nitrate(V)
I. To about 5cm3 of a salt solution in a test tube add four drops of
Lead(II)nitrate(V)solution. Preserve.
Observation Inference
White precipitate/ppt SO42-
, SO32-
, CO32-
, Cl- ions
II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid .
Preserve.
Observation 1
Observation Inference
White precipitate/ppt persists SO42-
, Cl- ions
Observation 2
Observation Inference
White precipitate/ppt dissolves SO32-
, CO32-
, ions
III.(a)To the preserved sample observation 1 in (II) above, Heat to boil.
45
All high school revision materials are available on www.kusoma.co.ke
Observation 1
Observation Inference
White precipitate/ppt persists on boiling SO42-
ions
Observation 2
Observation Inference
White precipitate/ppt dissolves on boiling Cl - ions
.(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified
potassium manganate(VII) /dichromate(VI).
Observation 1
Observation Inference
(i)acidified potassium manganate(VII)decolorized
(ii)Orange colour of acidified potassium
dichromate(VI) turns to green
SO32-
ions
Observation 2
Observation Inference
(i)acidified potassium manganate(VII) not
decolorized
(ii)Orange colour of acidified potassium
dichromate(VI) does not turns to green
CO32-
ions
Experiments/Observations:
(b)Using Barium(II)nitrate(V)/ Barium(II)chloride
I. To about 5cm3 of a salt solution in a test tube add four drops of Barium(II)
nitrate (V) / Barium(II)chloride. Preserve.
Observation Inference
White precipitate/ppt SO42-
, SO32-
, CO32-
ions
II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid .
Preserve.
Observation 1
Observation Inference
White precipitate/ppt persists SO42-
, ions
46
All high school revision materials are available on www.kusoma.co.ke
Observation 2
Observation Inference
White precipitate/ppt dissolves SO32-
, CO32-
, ions
III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified
potassium manganate(VII) /dichromate(VI).
Observation 1
Observation Inference
(i)acidified potassium manganate(VII)decolorized
(ii)Orange colour of acidified potassium
dichromate(VI) turns to green
SO32-
ions
Observation 2
Observation Inference
(i)acidified potassium manganate(VII) not
decolorized
(ii)Orange colour of acidified potassium
dichromate(VI) does not turns to green
CO32-
ions
Explanations
Using Lead(II)nitrate(V)
(i)Lead(II)nitrate(V) solution reacts with chlorides(Cl-), Sulphate (VI) salts
(SO42-
), Sulphate (IV)salts (SO32-
) and carbonates(CO32-
) to form the insoluble
white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate
(IV) and Lead(II)carbonate(IV).
Chemical/ionic equation:
Pb2+
(aq) + Cl- (aq) -> PbCl2(s)
Pb2+
(aq) + SO42+
(aq) -> PbSO4 (s)
Pb2+
(aq) + SO32+
(aq) -> PbSO3 (s)
Pb2+
(aq) + CO32+
(aq) -> PbCO3 (s)
(ii)When the insoluble precipitates are acidified with nitric(V) acid,
47
All high school revision materials are available on www.kusoma.co.ke
- Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and
thus their white precipitates remain/ persists.
- Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to
form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out
sulphur(IV)oxide and carbon(IV)oxide gases respectively.
. Chemical/ionic equation:
PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb
2+(aq) + SO2 (g)
PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb
2+(aq) + CO2 (g)
(iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed;
- Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on
cooling)
- Lead(II)sulphate(VI) do not dissolve in hot water thus its white
precipitate persists/remains on heating/boiling.
(iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced;
- sulphur(IV)oxide will decolorize acidified potassium manganate(VII)
and / or Orange colour of acidified potassium dichromate(VI) will turns to
green. Carbon(IV)oxide will not.
Chemical equation:
5SO32-
(aq) + 2MnO4
- (aq) +6H+(aq) -> 5SO4
2-(aq) + 2Mn
2+(aq) + 3H2O(l)
(purple) (colourless)
3SO32-
(aq) + Cr2O7
2-(aq) +8H+(aq) -> 3SO4
2-(aq) + 2Cr
3+(aq) + 4H2O(l)
(Orange) (green)
- Carbon(IV)oxide forms an insoluble white precipitate of calcium
carbonate if three drops of lime water are added into the reaction test tube when
effervescence is taking place. Sulphur(IV)oxide will not.
Chemical equation:
Ca(OH)2(aq)
+ CO2 (g) -> CaCO3(s) + H2O(l)
These tests should be done immediately after acidifying to ensure the gases
produced react with the oxidizing agents/lime water.
Using Barium(II)nitrate(V)/ Barium(II)Chloride
48
All high school revision materials are available on www.kusoma.co.ke
(i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate
(VI) salts (SO42-
), Sulphate (IV)salts (SO32-
) and carbonates(CO32-
) to form the
insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV)
and Barium(II)carbonate(IV).
Chemical/ionic equation:
Ba2+
(aq) + SO42+
(aq) -> BaSO4 (s)
Ba2+
(aq) + SO32+
(aq) -> BaSO3 (s)
Ba2+
(aq) + CO32+
(aq) -> BaCO3 (s)
(ii)When the insoluble precipitates are acidified with nitric(V) acid,
- Barium (II)sulphate(VI) do not react with the acid and thus its white
precipitates remain/ persists.
- Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the
acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/
bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.
. Chemical/ionic equation:
BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba
2+(aq) + SO2 (g)
BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba
2+(aq) + CO2 (g)
(iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced;
- sulphur(IV)oxide will decolorize acidified potassium manganate(VII)
and / or Orange colour of acidified potassium dichromate(VI) will turns to
green. Carbon(IV)oxide will not.
Chemical equation:
5SO32-
(aq) + 2MnO4
- (aq) +6H+(aq) -> 5SO4
2-(aq) + 2Mn
2+(aq) + 3H2O(l)
(purple) (colourless)
3SO32-
(aq) + Cr2O7
2-(aq) +8H+(aq) -> 3SO4
2-(aq) + 2Cr
3+(aq) + 4H2O(l)
(Orange) (green)
- Carbon(IV)oxide forms an insoluble white precipitate of calcium
carbonate if three drops of lime water are added into the reaction test tube when
effervescence is taking place. Sulphur(IV)oxide will not.
Chemical equation:
Ca(OH)2(aq)
+ CO2 (g) -> CaCO3(s) + H2O(l)
49
All high school revision materials are available on www.kusoma.co.ke
These tests should be done immediately after acidifying to ensure the gases
produced react with the oxidizing agents/lime water.
Summary test for Sulphate (VI) (SO42-
)and Sulphate(IV) (SO32-
) salts
Practice revision question
1. Study the flow chart below and use it to answer the questions that follow
Unknown salt
Lead(II)nitrate(V)
White precipitates of Cl-, SO4
2- ,
SO32-
and CO32-
Dilute nitric(V) acid
white ppt dissolves in
SO32-
and CO32-
white ppt persist /remains
in SO32-
and CO32-
Acidified KMnO4
decolorized in SO32-
White ppt with
lime water in CO32-
Acidified KMnO4
K2Cr2O7 / Lime water
White ppt dissolves
on heating in Cl-
White ppt persist
on heating in SO42-
in CO32-
Heat to boil
Sodium salt solution
White precipitate
Barium nitrate(VI)
(VI)(aq)
Colourless solution B
Gas G and colour of solution
changes orange to green
Acidified K2Cr2O7 Dilute HCl
50
All high school revision materials are available on www.kusoma.co.ke
(a)Identify the:
I: Sodium salt solution
Sodium sulphate(IV)/Na2SO3
II: White precipitate
Barium sulphate(IV)/BaSO3
III: Gas G
Sulphur (IV)Oxide /SO2
IV: Colourless solution H
Barium chloride /BaCl2
(b)Write an ionic equation for the formation of:
I.White precipitate
Ionic equation Ba2+
(aq) + SO32-
(aq) -> BaSO3(s)
II.Gas G
Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba
2+(aq)
III. Green solution from the orange solution
3SO32-
(aq) + Cr2O7
2-(aq) +8H+(aq) -> 3SO4
2-(aq) + 2Cr
3+(aq) + 4H2O(l)
(Orange) (green)
2. Study the flow chart below and answer the questions that follow.
51
All high school revision materials are available on www.kusoma.co.ke
(i)Write equation for the reaction taking place at:
I.The roasting furnace (1mk)
2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g)
II.The absorption tower (1mk)
H2SO4 (l) + SO3 (g) -> H2S2O7(l)
III.The diluter (1mk)
H2S2O7(l) + H2 O(l) -> 2H2SO4 (l)
(ii)The reaction taking place in chamber K is
SO2 (g) + 1/2O2 (g) SO3 (g)
I. Explain why it is necessary to use excess air in chamber K
To ensure all the SO2 reacts
II.Name another substance used in chamber K
Vanadium(V)oxide
3.(a)Describe a chemical test that can be used to differentiate between sodium
sulphate (IV) and sodium sulphate (VI).
Add acidified Barium nitrate(V)/chloride.
White precipitate formed with sodium sulphate (VI)
52
All high school revision materials are available on www.kusoma.co.ke
No white precipitate formed with sodium sulphate (IV)
(b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is
reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas
at s.t.p =22.4dm3)
Chemical equation
Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g)
Mole ratio Cu(s: SO2 (g) = 1:1
Method 1
1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3
(120 x 1000) g -> (120 x 1000) g x 22.4.dm3)
63.5 g
= 42330.7087
Method 2
Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles
63.5
Moles SO2 = Moles of Cu = 1889.7639 moles
Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4)
= 42330.7114
Green
solution T
White precipitate RAcidified
K2Cr2O7
Barium nitrate(V)
Colourless gas
V
Dilute nitric
(V) acid
4.Use the reaction scheme below to answer the
questions that follow.
(a)Identify the:
53
All high school revision materials are available on www.kusoma.co.ke
(i)cation responsible for the green solution T
Cr3+
(ii)possible anions present in white precipitate R
CO32-
, SO32-
, SO42-
(b)Name gas V
Sulphur (IV)oxide
(c)Write a possible ionic equation for the formation of white precipitate R.
Ba2+
(aq) + CO32-
(aq) -> BaCO3(s)
Ba2+
(aq) + SO32-
(aq) -> BaSO3(s)
Ba2+
(aq) + SO42-
(aq) -> BaSO4(s)
********************END**********************
This document was downloaded from www.kusoma.co.ke
The other documents available on this website include;
All secondary school schemes of work.
Form 1-form 4 revision papers in all subjects.
Marking schemes to most of the revision papers.
K.C.S.E past papers.
University past papers.
K.A.S.N.E.B past papers e.g. C.P.A
54
All high school revision materials are available on www.kusoma.co.ke
Primary school schemes of work.
Primary school revision papers to all classes.
Marking schemes to primary school past papers.
K.C.P.E past papers.